Answer
$C.-12.8cm$
Work Step by Step
We know that
$f_{in \space fluid}=[\frac{(n_{lens}-1)n_{fluid}}{n_{lens}-n_{fluid}}]f_{in\space air}$
We plug in the known values to obtain:
$f_{in\space fluid}=[\frac{(1.31-1)\times 1.5}{1.31-1.5}]\times 5.25cm=-12.8cm$
Thus the correct answer is option $C$, that is $-12.8cm$
