$(a)$ The situation is represented in the given figure. $O$ is the mid-point of line $AB.$

Distance between the two charges, $AB =20\, cm$

$\therefore AO = OB =10 \,cm$

Net electric field at point $O = E$

Electric field at point $O$ caused by $+3 \,\mu\, C$ charge,

$E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 \times 10^{-6}}{(O A)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1} \quad$ along $O B$

Where, $\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,N\,m ^{2} \,C ^{-2}$

Therefore, Magnitude of electric field at point $O$ caused by $-3\, \mu \,C$ charge,

$E_{2}=\left|\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{-3 \times 10^{-6}}{(O B)^{2}}\right|$$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1}$ along $O B$

$\therefore E=E_{1}+E_{2}$$=2 \times \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1} \quad$ along $O B$

$\therefore E=2 \times 9 \times 10^{9} \times \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} \,N\, C^{-1}$

$=5.4 \times 10^{6} \,N\,C ^{-1}$ along $OB$

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

$(b)$ A test charge of amount $1.5 \times 10^{-9} C$ is placed at mid-point $O$. $q=1.5 \times 10^{-9} \,C$

Force experienced by the test charge $= F$ $\therefore F = qE$

$=1.5 \times 10^{-9} \times 5.4 \times 10^{6}$

$=8.1 \times 10^{-3} \,N$

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 towards point $A$. Therefore, the force experienced by the test charge is $8.1 \times 10^{-3}\, N$ along $OA$.