We know that, $\alpha$ - decay formula is -

$$ { }_Z^A X \stackrel{-\alpha}{\rightarrow}{ }_{Z-2}^{A-4} Y $$

And the formula for $\beta$ is -

$$ { }_Z^A X \stackrel{-\beta}{\rightarrow}{ }_{Z-1}^A Y $$

Now for the given above question we see that, -

$$ \begin{aligned} &{ }_{92}^{238} M \stackrel{-2 \alpha}{\rightarrow}{ }_{92-4}^{238-8} N \\\\ &{ }_Y^X X \stackrel{-2 \beta^{+}}{\rightarrow}{ }_{86}^{230} L \end{aligned} $$

Therefore we can now calculate the number of neutrons in $L$, -

$$ L=230-86 $$

$$ L=144 $$

So, as we can see the no. of neutrons present in $\mathrm{L}$ is 144 .