Finding the Sum of series $S = \sum_{n=1}^{243} \frac{1}{n^{4/5}} $.

You're getting a very loose approximation with the integral if you start at $0$, because $\int_0^1 x^{-4/5}\,dx = 5$. That's the majority of the error.

To avoid this, write $$S = 1 + \sum_{n=2}^{243} \frac1{n^{4/5}}$$ and then use integrals to put upper and lower bounds on the sum from $2$ to $243$. Since the function $f(x) = \frac1{x^{4/5}}$ is decreasing for $x>0$, we have $$\int_2^{244} \frac1{x^{4/5}}\,dx < \sum_{n=2}^{243} \frac1{n^{4/5}} < \int_1^{243} \frac1{x^{4/5}}\,dx$$ and that should be enough to estimate the sum closely enough to find the floor $\lfloor S\rfloor$.

You will not find an exact value. You are probably expected to find an approximation.

You could use the trapezium rule to approximate the integral

$$\int_1^{243}f(x)dx \approx \frac12(f_1+f_2)+\frac12(f_2+f_3)+...\frac12(f_{241}+f_{242})+\frac12(f_{242}+f_{243})$$ $$= -\frac12(f_1+f_{243})+\sum_1^{243}f_n $$ Then $$I=\int_1^{243}x^{-4/5}dx=[5x^{1/5}]_1^{243}=5(3-1)=10$$ Therefore $$\sum_1^{243}\frac{1}{n^{4/5}} \approx I+\frac12(\frac{1}{1^{4/5}}+\frac{1}{243^{4/5}})=10+\frac12(1+\frac{1}{81}) =10\frac{41}{81}\approx 10.5062$$

You could make this more exact using the error term for the trapezium rule which is $-M(b-a)/12n^2=-M/12n$ where $M$ is the average value of $f''(x)$ over the interval.