A similar problem which is used to solve the given question is given in the following link :
Question No. 9 here.
The solution is explained for the given problem by TA as :
Black colored surface : $x\%$
White colored surface : $(100-x)\% = y\% = y/100$
So, $x+y=100$
Now, probability that a particular vertex has black color is = $\frac{x}{100}$
and probability that a particular vertex has white color is = $\frac{y}{100}$
Now, define a random variable $X_i$ as :
$X_i$ = $\left\{\begin{matrix} 1 & if \;i^{th} \; vertex\; is \;colored\; white \\ 0& otherwise \end{matrix}\right.$
Now, $P(X_i=1) = \frac{y}{100}$ where $1\leqslant i \leqslant 8$
Now, consider a new random variable $X=$ $\sum_{i=1}^{8}X_i$.
So, Expected no. of vertices of color white = $E(X)= E(\sum_{i-1}^{8}X_i) = \frac{y}{100}*8$ (Using linearity of expectation)
Since, all its '8' vertices are white which touches the sphere, so expected no. of vertices of color white should be strictly greater than 7
So, $\frac{8y}{100} > 7$
$ \Rightarrow y>700/8$
$ \Rightarrow 100-x>700/8$
$ \Rightarrow x>100/8$
$ \Rightarrow x<12.5 \%$
similar link.
Now, the same concept can be applicable for the given question.
Since, here the eight vertices of the cube touching only red colored parts of the surface of the sphere.
So, here also $y > \frac{700}{8}\%$
$\Rightarrow y > \frac{7}{8}$
So, It is saying that we need 7/8 of the total surface area of sphere to be red, so that all the $8$ vertices of the cube touch the sphere.
Since, radius of the sphere is $1$ metre.
So, $\frac{7}{8}$ of the total surface area of sphere = $\frac{7}{8}*4\pi(1)^2=10.99$ $metre^2$
So, area of the red part should be greater than $10.99$ $metre^2$
So, I think answer should be (D).
same question is asked here