Contrary to @whuber's claim, the mean of x and y are contained in the information given.
Okay, so you have the line equation
$$y_i=\alpha +x_i\beta + e_i$$
estimates $\hat{\beta}=r\frac{s_y}{s_x}$ and $\hat{\alpha}=\overline{y}-\hat{\beta}\overline{x}$.
where $r$ is the correlation. The question doesn't state whether the standard deviation (0.482) is for $s_y$ or $s_x$ (the MLE standard deviation, with divisor $n$). Either way, you can work out the either from the info given. for their ratio must satisfy:
$$\frac{\hat{\beta}}{r}=\frac{s_y}{s_x}$$
The slope can't be negative if the correlation is positive, so I have assumed that you have done something incorrectly (for you have correlation of 0.117, and slope of -0.00024; this is impossible). This will affect the numbers, but not the general method. So I will assume the standard deviations are both known, but not write in the specific values. The same goes for the rest of the actual numbers.
Now the variance of $\hat{\beta}$ is given by:
$$var(\hat{\beta})=s_e^2(X^TX)^{-1}_{22}=\frac{s_e^2 (X^TX)_{11}}{|X^TX|}$$
Note that $(X^TX)_{11}=n$ and $s_e^2$ is the "mean square error". The variance of $\alpha$ is given by:
$$var(\hat{\alpha})=s_e^2(X^TX)^{-1}_{11}=\frac{s_e^2 (X^TX)_{22}}{|X^TX|}$$
Now $(X^TX)_{22}=\sum_i x_i^2 = n(s_x^2+n\overline{x}^2)$
And dividing these two variances gives:
$$\frac{var(\hat{\alpha})}{var(\hat{\beta})}=\frac{(X^TX)_{22}}{(X^TX)_{11}}=\frac{n(s_x^2+n\overline{x}^2)}{n}=s_x^2+n\overline{x}^2$$
Now all quantities in the equation are known, except for the mean $\overline{x}$. So we can re-arrange this equation and solve for the mean:
$$\overline{x}=\pm\sqrt{\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}}$$
But we know from the start that $x_i>0$ - you can't drive "negative miles". So only the positive square root is to be taken. The rest is straight-forward CI stuff. The estimate of the mean $\hat{\overline{y}}$ is given by:
$$\hat{\overline{y}}=\hat{\alpha}+\hat{\beta}\overline{x}=\hat{\alpha}+\hat{\beta}\sqrt{\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}}=\overline{y}$$
And the variance is given by:
$$var(\hat{\overline{y}})=var(\hat{\alpha})+\overline{x}^2 var(\hat{\beta})+2\overline{x}cov(\hat{\alpha},\hat{\beta})$$
Now the covariance is equal to:
$$cov(\hat{\alpha},\hat{\beta})=s_e^2(X^TX)^{-1}_{21}=-\frac{s_e^2 (X^TX)_{21}}{|X^TX|}=-\frac{s_e^2 n\overline{x}}{ns_x^2}=-\frac{s_e^2 \overline{x}}{s_x^2}$$
And so the variance is given by:
$$var(\hat{\overline{y}})=var(\hat{\alpha})+\overline{x}^2 var(\hat{\beta})-2\frac{s_e^2 \overline{x}^2}{s_x^2}=var(\hat{\alpha})+\frac{\frac{var(\hat{\alpha})}{var(\hat{\beta})}-s_x^2}{n}\left(var(\hat{\beta})-2\frac{s_e^2}{s_x^2}\right)$$
So you construct your $100(1-P)$% confidence interval by choosing $T_{1-P/2}^{(n-2)}$ as the $P/2$ quantile of standard T distribution with $n-1$ degrees of freedom (which effectively equal to the standard normal, as $n-1=100$), and you have:
$$CI=\overline{y}\pm T_{1-P/2}^{(n-2)}\sqrt{var(\hat{\overline{y}})}$$
And all quantities are calculable, given the information.
