Consider the following nuclear reactions $$_{92}^{238}M \rightarrow {}_y^x N + {}_2^4 He$$; $$_y^N N \rightarrow {}_B^{A} L + 2\beta^+$$. The number of neutrons in the element $$L$$ is

The first nuclear reaction is $$^{238}{92}M \rightarrow {}^{x}{y}N + {}^{4}_{2}He$$.

Balancing the mass numbers, $$238=x+4$$, therefore $$x=234$$.

Balancing the atomic numbers, $$92=y+2$$, therefore $$y=90$$.

Hence, $$^{238}{92}M \rightarrow {}^{234}{90}N + {}^{4}_{2}He$$.

The second reaction is $$^{234}{90}N \rightarrow {}^{A}{B}L + 2\left(^{0}_{+1}e\right)$$.

Balancing the mass numbers, $$234=A+2(0)$$, therefore $$A=234$$.

Balancing the atomic numbers, $$90=B+2(+1)$$, therefore $$B=88$$.

Hence, $$^{234}{90}N \rightarrow {}^{234}{88}L + 2\left(^{0}_{+1}e\right)$$.

The number of neutrons in element $$L$$ is

$$\text{Number of neutrons}=A-B=234-88=146$$

Therefore, the number of neutrons present in element $$L$$ is $$\boxed{146}$$.

Hence, the correct answer is Option (B).

Note: If the answer key shows $$144$$, it is incorrect. The value $$144$$ corresponds to the number of neutrons in the intermediate nucleus $$N$$:

$$234-90=144$$