Quite close to Ramanujan!

$x^{300}<1728^{200}$ $x^{300}<12^{600}$ $x<12^{2}$ $x<144$ Therefore the greatest possible integral value of $x$ is $143$ .

${ x }^{ 300 }<{ 1728 }^{ 200 }\\ { x }^{ 3 }<{ 1728 }^{ 2 }\\ { x }^{ 3 }<({ { 2 }^{ 6 }\cdot { 3 }^{ 3 }) }^{ 2 }\\ x<{ 2 }^{ 4 }\cdot { 3 }^{ 2 }\\ x<144\\ x=143$

$\begin{aligned} x^{300} < 1728^{200}\\ \implies {{\sqrt{x}}^{2}}^{300} < {{12}^3}^{200}\\ \implies {\sqrt{x}}^{600} < {12}^{600}\\ \implies \sqrt{x} < 12\\ \text{Squaring on both sides}:-\\ \implies x < 144 \end{aligned}$

$\text{So, the greatest possible integral value of x} = (144-1) = \boxed{143}$

x ^ 300 < 1728 ^ 200 300 * log ( x ) < 200 * log ( 1728 ) 3 log x < 2 log 1728 now you can binary search for the answer.

int l = 1;

int r = 1000;

double res = log(1728) * 2;

while ( l < r ) {

    int mid = ( l + r ) >> 1 ;

    double temp = log(mid) * 3;

    if( temp < res ){

        l = mid;

    }

    else{

        r = mid - 1;

    }

}

cout << l;

Note that $1728^{200}$ is $(12^{3})^{200} = 12^{600}$ and not $12^{3^{200}}$ , which is something ridiculous.

$x^{300} < 1728^{200} = 12^{600}$

Take the 300th root.

$x < 12^{600} \times 12^{\frac{1}{300}}= 12^{\frac{600}{300}} = 12^{2} = 144$

Next smallest integer $x$ is 143 .

X^300<(12^2)^300,x^300<144^300,so x.max=143####

X^300 < 1728^200 X^3 < 1728^2 X^3 < (2^4 . 3^2)^3 X < 2^4 . 3^2 X < 144

So the greatest possible...integral value is 143