1. Let $c=\cos\theta.$ Then the given complex number becomes $z=\frac{3+2ic}{1-3ic}.$ We are given that the real part of $z$ is zero.

2. Rationalize the denominator: z=\frac{3+2ic}{1-3ic}\cdot\frac{1+3ic}{1+3ic}=rac{(3+2ic)(1+3ic)}{1+9c^2}.

Now expand the numerator: $(3+2ic)(1+3ic)=3+9ic+2ic+6i^2c^2=3+11ic-6c^2,$ because $i^2=-1$.

So, $z=\frac{3-6c^2+11ic}{1+9c^2}.$

3. Therefore, $\Re(z)=\frac{3-6c^2}{1+9c^2}.$ Given that the real part is zero, $\frac{3-6c^2}{1+9c^2}=0.$ Since $1+9c^2>0$, we must have $3-6c^2=0$ $6c^2=3$ $c^2=\frac12.$ Thus, $\cos^2\theta=\frac12.$

Since $\theta\in\left(0,\frac\pi2\right)$, this gives $\sin^2\theta=1-\cos^2\theta=1-\frac12=\frac12.$

4. Now compute the required value: $\sin^2\theta+\cos^2\theta=\frac12+\frac12=1.$

Hence the required integer is $\boxed{1}.$