It would be possible if the mean is the mean of a binary variable coded using the values 0 or 1. In that case, the mean would actually be the proportion of 1s. In that case the odds of a 1 would be $\frac{0.7}{1-0.7}\approx 2.3$, i.e. within the control group we expect to find 2.3 1s for ever 0. The odds of a 1 for the test group would be $\frac{0.015}{1-0.015}\approx 0.015$. The odds ratio would than be just the ratio of those odds: $\frac{\frac{.7}{1-.7}}{\frac{0.015}{1-0.015}} = \frac{0.7}{1-0.7} \times \frac{1-0.015}{0.015}= \frac{0.7(1-0.015)}{(1-0.7)0.015}\approx153$, that is the odds of a 1 in the control group is a 153 times larger than the odds of a 1 in the test group.
However, your means are not proportions, so this method won't work for you. I know this because if the mean for the control group were a proportion, then the standard deviation would have been $\sqrt{0.7*(1-0.7)}\approx0.46$, which is completely different from the standard deviation you report in the table.
If you want to find out if there is anything possible, then we need to know more about your problem: In particular what is your dependent variable and how is it measured?