Alkane free radical halogenation
This question is based on relative rates of reaction of different $\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\ce{3^\circ : 2^\circ : 1^\circ}$ $\ce{H}$ is in order $5:3.8: 1$.

Using this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply number of identical hydrogen and its rate of reaction with chlorine.

So there is only $1$ $3^\circ$ $\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\circ$ $\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.

Refer: Wikipedia