In a binomial distribution $$B\left(n, p = \frac{1}{4}\right)$$, if the probability of at least one success is greater than or equal to $$\frac{9}{10}$$, then $$n$$ is greater than

The probability of getting at least one success in a binomial distribution is obtained by subtracting the probability of getting no success from 1.

Here $$p = \frac14$$, so the probability of failure in a single trial is $$q = 1-p = \frac34$$.
Probability of zero successes in $$n$$ trials is $$q^{\,n} = \left(\frac34\right)^n$$.

Given condition:
$$1 - \left(\frac34\right)^n \;\ge\; \frac{9}{10}$$

Rearrange:

$$\left(\frac34\right)^n \;\le\; \frac1{10} \quad -(1)$$

Take common (base-10) logarithms on both sides:

$$n \,\log_{10}\!\left(\frac34\right) \;\le\; \log_{10}\!\left(\frac1{10}\right) = -1 \quad -(2)$$

Since $$\log_{10}\!\left(\frac34\right)$$ is negative, dividing by it reverses the inequality:

$$n \;\ge\; \frac{-1}{\log_{10}\!\left(\frac34\right)}$$

Write the logarithm difference explicitly:

$$\log_{10}\!\left(\frac34\right) = \log_{10} 3 \;-\; \log_{10} 4$$

Thus

$$n \;\ge\; \frac{-1}{\log_{10} 3 - \log_{10} 4} \;=\; \frac{1}{\log_{10} 4 - \log_{10} 3}$$

Hence $$n$$ must be greater than $$\dfrac{1}{\log_{10} 4 - \log_{10} 3}$$.

Option A which is: $$\frac{1}{\log_{10} 4 - \log_{10} 3}$$