The standard followed by IEEE-754 for single precision is:
$\begin{array}{|l|l|l|} \hline \text { Sign bit (1 bit) } & \text { Exponent bit (8 bit) } & \text { Mantissa (23 bit) } \\ \hline \end{array}$
which is the same as what is mentioned in the question.
IEEE-754 follows implicit normalization where ($E$ is the biased exponent) $E \neq 00 \ldots 0$ and $E \neq 111 \ldots 1$, i.e., exponents all zero's and all 1's are not allowed while the mantissa can be anything ($M = X X \ldots X$).
Now, we are asked to find the largest finite positive value so:
$S=0$ (for positive)
$\mathrm{E}=11111110$ (as all 1's not allowed, i.e., 254 in decimal)
$\mathrm{M}=1111 \ldots 1$ (all 1's as no constraint on the mantissa part)
Let $e$ be the original exponent.
$\mathrm{E}$ (biased exponent) $= e +$ bias (for single precision in IEEE-754 bias $= 127$)
so, $e = 254 - 127 = 127$
Value $= (-1)^0 \times (1.11111111111111111111111) \times 2^{127}$
$= 111111111111111111111111 \times 2^{-23} \times 2^{127}$
$= (2^{24} - 1) \times 2^{-23} \times 2^{127}$
$= (2 - 2^{-23}) \times 2^{127}$
$= (1 + (1 - 2^{-23})) \times 2^{127}$
Hence, Option B is correct.