While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for the what the variance is actually "doing." An equivalent expression for the variance is in this case is:

$Var(X)$ =$ \sum_{i=1}^6(X_i-\bar{X})^2\over{6}$.

Now, you know the mean, $\bar{X}=3.5$, so you simply need to take the die's $i$th's face value $i=1, 2, . . . , 6$, $X_i$ and subtract it from the mean, square it, and divide it by 6. In effect this gives you an average of how far away each die value is from its mean. So $Var(X)$ is given by:

${(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}\over{6}$= $17.5\over{6}$=$105/36$, the same answer @dsaxton provided.

We square the values of $X_i-\bar{X}$ because if we don't, then the sum of the values will add to zero and the negative numbers will cancel out the positive numbers.

While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for the what the variance is actually "doing." An equivalent expression for the variance in this case is:

$Var(X)$ =$ \sum_{i=1}^6(X_i-\bar{X})^2\over{6}$.

Now, you know the mean, $\bar{X}=3.5$, so you simply need to take the die's $i$th's face value $i=1, 2, . . . , 6$, $X_i$ and subtract it from the mean, square it, and divide it by 6. In effect this gives you an average of how far away each die value is from its mean. So $Var(X)$ is given by:

${(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}\over{6}$= $17.5\over{6}$=$105/36$, the same answer @dsaxton provided.

We square the values of $X_i-\bar{X}$ because if we don't, then the sum of the values will add to zero and the negative numbers will cancel out the positive numbers.