Applying Bragg’s equation

$n\lambda=2d\sin\theta$

Giving $n=1,\lambda=1.54A^{\large\circ},\theta=22.5^{\large\circ}$

Using relation $n\lambda=2d\sin \theta$

$d=\large\frac{1.54}{2\sin 22.5^{\large\circ}}=\frac{1.54}{2\times 0.383}$$=2.01A^{\large\circ}$

Hence (B) is the correct answer.