since, no external force is acting on the system, we can apply conservation of linear momentum.

Speed of $200 \mathrm{kg}$ trolley $=\frac{36\times 1000}{60\times 60}=10 \mathrm{m} {\mathrm{s}}^{-1}$

If $u$ be the initial velocity of trolley, $v_b$ be the absolute velocity of the boy after the beginning of journey of the boy, their relative velocity is $4$.

So, $v^{\text{'}}-v_b=4\Rightarrow v_b=\left(v^{\text{'}}-4\right)$

Now, applying law of conservation of momentum,

Momentum before the boy begins tu run
$=$ Momentum after the beginning of boy's running

$\Rightarrow 220\times 10=200v^{\text{'}}+20\left(v^{\text{'}}-4\right)$

$2200=220v^{\text{'}}-80\Rightarrow 220v^{\text{'}}=2280$

$v^{\text{'}}=\frac{2280}{220}=10.36 \mathrm{m} {\mathrm{s}}^{-1}$