Answer
$(D)~2.29eV$
Work Step by Step
We know that
$W_{\circ}=\frac{K_2-K_1\frac{\lambda_1}{\lambda_2}}{\frac{\lambda_1}{\lambda_2}-1}$
We plug in the known values to obtain:
$W_{\circ}=\frac{2.57eV-(0.550eV)(\frac{433.9nm}{253.5nm})}{(\frac{433.9nm}{253.5nm})-1}$
$W_{\circ}=2.29eV$
