$\mathrm{LiH} + {\mathrm{B}}_2{\mathrm{H}}_6 \stackrel{\mathrm{dry} \mathrm{ether}}{\to } \mathrm{X}$

In this reaction 'X' is used as

The reaction is given below:

$2\mathrm{LiH} + {\mathrm{B}}_2{\mathrm{H}}_6 \stackrel{\mathrm{dry} \mathrm{ether}}{\to } 2{\mathrm{LiBH}}_4$

$2{\mathrm{LiBH}}_4$ is lithium borohydride and it is a reducing agent.

Therefore, the correct option is (B).