It is given that we have unit length stick, so let the stick be broken at point $(x,0)$ supposing that stick is placed horizontally from origin $(0,0)$. Such that one part would be having length of $x$ units and another would be having length of $1-x$ units. We are instructed to further break the largest part of stick in two pieces.
Let, $x$ length be the largest part, such that it would be broken into $\frac{3x}{7}$ and $\frac{4x}{7}$ unit lengths. Now we have stick broken into three pieces of length given in $3$-ary tuple $(1-x, \frac{3x}{7}, \frac{4x}{7})$.
If $a$, $b$, and $c$ are lengths of the sides of the triangle, with no side being greater than $c$, then the triangle inequality states that $a+b\ge c$.
$\underline{\textbf{Observations}}$
To satisfy triangle inequality, no side of the triangle should be greater than $\frac{1}{2}$ unit length. (Why?)
So, length of any side of the triangle should lie in $(0,\frac{1}{2})$.
Keeping the above observation in mind, we can say that length of any side of triangle should be in $[\frac{1}{2}, 1]$. (Since, we need to find for the case when triangle CANNOT be formed).
We want the values of $x$ for which the triangle inequality is not satisfied. So, we will find such values in cases.
$\textbf{Case 1:} \ (1-x) + \frac{3x}{7} < \frac{4x}{7} \implies x>\frac{7}{8}$ (in $(\frac{7}{8}, 1])$
$\textbf{Case 2:} \ (1-x) + \frac{4x}{7} < \frac{3x}{7} \implies x>\frac{7}{6}$ (Not Possible, Out of Domain)
$\textbf{Case 3:} \ \frac{3x}{7} + \frac{4x}{7} < (1-x) \implies x<\frac{1}{2}$ (Not Possible, Out of Domain)
$$P(x>\frac{7}{8}) = 1-P(x \le \frac{7}{8}) = 1- \int_{\frac{1}{2}}^{\frac{7}{8}}\underbrace{\dfrac{1}{1-\frac{1}{2}}}_{pdf}dx = \dfrac{1}{4}$$
$\textbf{Option (A) is correct}$
