​$$\sqrt{201\frac{49}{100000}+49\frac{1}{10000}-\frac{5}{1000}+6\frac{45}{10000}}=x$$​
Find the approximate value of $$x^2$$​ ?

Correct option is B

Given:
Expression: $$\sqrt{201\frac{49}{100000}+49\frac{1}{10000}-\frac{5}{1000}+6\frac{45}{10000}}=x$$​​
Solution:
​$$\sqrt{201\frac{49}{100000}+49\frac{1}{10000}-\frac{5}{1000}+6\frac{45}{10000}} \\ \ \\= \sqrt{\left(201+\frac{49}{100000}\right)+\left(49+\frac{1}{10000}\right)-\frac{5}{1000}+\left(6+\frac{45}{10000}\right)}\\ \ \\= \sqrt{201+49+6+\frac{49}{100000}+\frac{1}{10000}-\frac{5}{1000}+\frac{45}{10000}}\\ \ \\= \sqrt{256+\frac{49}{100000}+\frac{10}{100000}-\frac{500}{100000}+\frac{450}{100000}}\\ \ \\= \sqrt{256+\frac{49+10-500+450}{100000}} \\ \ \\= \sqrt{256+\frac{9}{100000}}\\ \ \\= \sqrt{256.00009}$$​​
Now,  

equating both side, 

​$$x^2 = \sqrt{256.00009} \\ \ \\ x^2 \approx 256$$​​
Final Answer
So the correct answer is (b)