This is the limit

Let $\displaystyle y = \lim_{n \to \infty} \frac{1}{n^{2m}}[(n^2 +1^2)(n^2 + 2^2).....(n^2 + n^2)]^{m/n}$

Taking logarithm both sides and take $n^{2m}$ inside the big braces, we get

$\displaystyle\log y = \lim_{n \to \infty} \frac{m}{n}\log [(\frac{n^2 +1}{n^2})(\frac{n^2+2^2}{n^2})....(\frac{n^2 + n^2}{n^2})]$

$\displaystyle\log y = \frac{m}{n} \sum_{r=1}^{n} \log(1 + \frac{r^2}{n^2})$

$\displaystyle\log y = m \int\limits_0^1 \log(1 + x^2) dx$ $\displaystyle \log y = m (\pi/2 + \log 2 -2)$

$\displaystyle y = e^{\log{2e^{\pi/2 - 2}}^m}$

$\displaystyle y = (2e^{\pi/2 -2})^m$

$a+b = \pi/2 -2 +2 = \pi/2$ which is the final answer.

$\begin{aligned} L & = \lim_{n \to \infty} \frac 1{n^{2m}} \left[(n^2+1^2)(n^2+2^2)(n^2+3^2)...(n^2+n^2)\right]^\frac mn \\ & = \exp \left(\ln \left(\frac 1{n^{2m}} \left[\prod_{r=1}^n (n^2+r^2)\right]^\frac mn \right) \right) & \small \color{#3D99F6}{\text{Note that }\exp(x) = e^x} \\ & = \exp \left(\frac mn \sum_{r=1}^n \ln (n^2+r^2) - 2m \ln(n) \right) \\ & = \exp \left(m \sum_{r=1}^n \frac { \ln (n^2+r^2) - 2\ln(n)}n \right) \\ & = \exp \left(m \left[ \ln(2) + \frac \pi 2 - 2 \right] \ \right) \\ & = \left(e^{\ln(2) + \frac \pi 2 -2} \right)^m \\ & = \left(2e^{\frac \pi 2 -2} \right)^m \end{aligned}$

$\implies b = \frac \pi 2 \approx \boxed{1.57}$