1. Let the magnitudes of the vectors be $|\vec P|=P,\quad |\vec Q|=Q,\quad |\vec R|=R$ with $\vec R=\vec P+\vec Q$ and given $R=P.$

2. Use the magnitude formula for the sum of two vectors. If the angle between $\vec P$ and $\vec Q$ is $\phi$, then $R^2=P^2+Q^2+2PQ\cos\phi.$ Since $R=P$, we get $P^2=P^2+Q^2+2PQ\cos\phi.$ Therefore, $Q^2+2PQ\cos\phi=0.$ So, 2P\cos\phi=-Q. \tag{1}

3. Now consider the new resultant $\vec S=2\vec P+\vec Q.$ But since $\vec R=\vec P+\vec Q,$ we can write $\vec S=\vec P+\vec R.$

4. We need the angle between $\vec S$ and $\vec Q$. Check their dot product: $\vec S\cdot \vec Q=(2\vec P+\vec Q)\cdot \vec Q=2\vec P\cdot \vec Q+Q^2.$ Using $\vec P\cdot \vec Q=PQ\cos\phi$, $\vec S\cdot \vec Q=2PQ\cos\phi+Q^2.$

5. From equation (1), $2P\cos\phi=-Q.$ Multiplying by $Q$, $2PQ\cos\phi=-Q^2.$ Hence, $\vec S\cdot \vec Q=-Q^2+Q^2=0.$

6. Since the dot product is zero, $\vec S$ is perpendicular to $\vec Q$. Therefore, the angle made by the resultant of $2\vec P$ and $\vec Q$ with $\vec Q$ is $\theta=90^circ.$

Thus, the required integer is: $\boxed{90}$