Answer
$5.9\mu m$
Work Step by Step
We know that
$2y=2Ltan [sin^{-1}(\frac{1.22\lambda_{vacuum}}{nD})]$
We plug in the known values to obtain:
$2y=2(25.4\times 10^{-3}m)tan[sin^{-1}(\frac{1.22(550\times 10^{-9}m)}{(1.36)(4.25\times 10^{-3}m)})]$
This simplifies to:
$2y=5.9\times 10^{-6}m=5.9\mu m$
