3

We have

$$\sqrt{\frac{1}{45a}-1}=2025a^2⟹\sqrt{\frac{1}{45a}-1}=(45^2)a^2$$

So we get

$$\frac{1}{45a}=45^4{a}^4+1⟹45^5{a}^5+45a-1=0$$

Notice that the polynomial is equal to

$$(45^2{a}^2-45a+1)(45^3{a}^3+45^2{a}^2-1)=0$$

However, the first quadratic polynomial

$$45^2{a}^2-45a+1=0$$

has no real number solution because

$$(-45{)}^2-4\times 45^2\times 1=-6075<0$$

So we know that the second quadratic polynomial

$$45^3{a}^3+45^2{a}^2-1=0⟹45^2{a}^2(45a+1)=1$$

Finally, we have

$$45a^3+a^2=a^2(45a+1)=\frac{1}{45^2}=\boxed{\frac{1}{2025}}$$