Volume Strength of $H_2O_2$ is the volumes of $O_2$ released on decomposition of one volume of $H_2O_2$ at STP.
On heating $H_2O_24 decomposes as follows
$2H_2O_2 \Rightarrow \Delta 2H_2O + O_2$
If we take 1 mole $H_2O_2$ solution and decompose it completely it will release 0.5 mole of O2. .
At STP one mole of any gas occupies 22.4L
So, 0.5 mole of $O_2$ will occupy 22.4L/mole $\times$ 0.5mole=11.2L.
If the molarity of a $H_2O_2$ solution is M, 1L of this solution will have M moles of $H_2O_2$ in it and 1L of it on decomposition will give $\large\frac{M}{2}$ moles of $O_2$. $\large\frac{M}{2}$ moles of $O_2$ will occupy 22.4L/mole $\times \large\frac{M}{2}$ mole=11.2ML. This 11.2M L of $O_2$ has come from 1L of $H_2O_2$ solution of molarity M.
So, if the molarity of a $H_2O_2$ solution is M, its volume strength will be 11.2M.
Hence (a) is the correct answer.