My, what a large number you are!

Surely the poser does not expect us to evaluate the RHS directly!

By grouping the powers appropriately write

$m^2=({10}\times{9}\times{5})^2(10\times{5})=450^2\times50$

and so

$m=450\times\sqrt{50}$

and so

$450\times\sqrt{49}<m<450\times\sqrt{64}$

and it follows that

${450\times7}<{m}<450\times8$

$\implies 3150<m<3600$

$\implies 3000<m<4000$

$\huge m^2 = 2^3 \times 3^4 \times 5^6$

$\huge m = \sqrt{8} \times 3^2 \times 5^3$

$\huge m = 2 \times \sqrt{2} \times 9 \times 125$

$\huge m = 2250 \times\sqrt{2}$

$\huge m = 2250 \times (1.414)$

$\huge m= 3181.5$

$\huge So , \boxed{3000 < m < 4000} :)$

$2^2 \times\ 3^4 \times\ 5^6 < m^2 < 2^4 \times\ 3^4 \times\ 5^6$ $\Rightarrow$ $2 \times\ 3^2 \times\ 5^3 < m < 2^2 \times\ 3^2 \times\ 5^3$ $\Rightarrow$ $2250 < m < 4500$ . Now $m = 2250\sqrt{2}\therefore\ 3000< m < 4000$

$m^2 = 2^3 \times 3^4 \times 5^6 \\ m^2 = 10125000\\ m = \sqrt10125000\\ m = 2250\sqrt2\\ 2250\sqrt2 \approx 3182 \\ \therefore 3000 < m < 4000$

Taking the square root

$m = 2 \times 9 \times 125 \times \sqrt{2} = (4 + 4 + 1) \times 250 \sqrt{2} = 2250\sqrt{2}$

Now, as $1 < \sqrt{2} < 1.5 \;\;\; \Rightarrow \;\;\; 2000 < 2250 < 2250\sqrt{2} = m < 3375 < 4000$ .

m^2 = 2^3x 3^4x5^6 = 10125000. Root of 19125000 approximately more than 3000. So m lies between3000 & 4000.

Easy way is to find (4000) ^2 and (3000)^2 Therefore, we can surely say that m must be between 3000 and 4000 as other inequalities do not satisfy the equation

Sqrt (10125000) = 3181.9805 {Correct to 4 decimal places.}