(a) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB =20 cm
$\therefore$ AO = OB = 10 cm
The net electric field at point O = E

The electric field at point O caused by + 3$\mathrm{\mu }$C charge,

 

Where, Eo = Permittivity of free space and $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$
Therefore,
The magnitude of the electric field at point O caused by - 3$\mathrm{\mu }$C charge,

Therefore, the electric field at mid-point O is $5.4\times {10}^6{\mathrm{NC}}^{-1}$ along OB.

 

(b) A test charge of amount $1.5\times {10}^{-9}C$ is placed at midpoint O.
$q=1.5\times {10}^{-9}C$
Force experienced by the test charge = F
$\begin{array}{l}\therefore \mathrm{F}=\mathrm{qE}\\ =1.5\times {10}^{-9}\times 5.4\times {10}^6\\ =8.1\times {10}^{-3}\mathrm{N}\end{array}$
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted to point A.
Therefore, the force experienced by the test charge is $8.1\times {10}^{-3}$ N along OA.