Answer
$E_m=1.03\frac{kV}{m}$
Work Step by Step
We know that;
$I=\frac{E_m^2}{2\mu_{\circ}c}$
This can be rearranged as:
$E_m=\sqrt{2I\mu_{\circ}c}$
We plug in the known values to obtain:
$E_m=\sqrt{2(1400)(4\times 3.1416\times 10^{-7})(3\times 10^8)}=1.03\times 10^3=1.03\frac{kV}{m}$