Two resistors 2 and 3 are connected in the gaps of bridge as shown...

In a Wheatstone bridge with 2Ω and 3Ω, the null point is found when $\frac{R_1}{R_2} = \frac{l}{L-l}$.
Initial balance: $\frac{2}{3} = \frac{l_1}{L-l_1}$ gives $l_1 = \frac{2L}{5}$.
For standard wire length $L = 100$ cm, $l_1 = 40$ cm.
When unknown resistance $R$ is connected in parallel with 3Ω, equivalent resistance becomes $\frac{3R}{3+R}$.
New null point: $\frac{2}{\frac{3R}{3+R}} = \frac{l_2}{L-l_2}$, which simplifies to $\frac{2(3+R)}{3R} = \frac{l_2}{L-l_2}$.
Null point shifts 22.5 cm toward Y: $l_2 = 40 + 22.5 = 62.5$ cm.
Substituting: $\frac{2(3+R)}{3R} = \frac{62.5}{37.5} = \frac{5}{3}$
$6(3+R) = 15R \Rightarrow 18 = 9R \Rightarrow R = 2$ Ω