Keep in mind you are allowed to perform basic math operations (+,–,×,÷) with the balanced equations of chemical reactions. I'd recommend to start with undesired component(s) ($\ce{W}$) which should not appear in final equation, and try to cancel them out.
Here, to obtain target reaction and eliminate $\ce{W}$, you need to subtract doubled first reaction from the second one (e.g. "$(2) - 2 \cdot (1)$"), which results in
\begin{align}
\require{cancel}
\ce{\cancel{\ce{2W}} + 3X - \cancel{\ce{2W}} - 2X &-> 2Z + 2Y - 4Y} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 \\
\ce{X + 2Y &-> 2Z} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1
\end{align}
$$\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 = \pu{-150 kcal} -2 \cdot (\pu{-200 kcal}) = \pu{250 kcal}$$