Answer
Please see the work below.
Work Step by Step
We know that
$\Delta U=\frac{1}{2}Kx_{\circ}^2$
This can be rearranged as:
$K=\frac{2\Delta V}{x_{\circ}^2}$
We plug in the known values to obtain:
$K=\frac{2\times 2\times 10^{-21}}{(1.46\times 10^{-12})^2}=1879\frac{N}{m}$
