Step 1: Understand the question.
Light of energy $4.2\ \text{eV}$ falls on a metal sphere of radius $10\ \text{cm}$ with work function $2.4\ \text{eV}$. Electrons fly off. We must find how many electrons leave before the emission stops.
Step 2: Why emission stops.
Each time an electron leaves, the sphere is left a bit more positive. This positive charge pulls the electrons back. After enough electrons leave, the sphere is positive enough to stop even the fastest electron. Emission stops at that point.
Step 3: Find the maximum kinetic energy.
By Einstein's photoelectric equation, the fastest electron has energy:
\[ K_{\max} = E - \phi = 4.2 - 2.4 = 1.8\ \text{eV} \]
Step 4: Find the stopping potential.
Emission stops when the sphere's potential reaches the value that just holds back a $1.8\ \text{eV}$ electron. That potential is $V = 1.8\ \text{V}$.
Step 5: Relate potential to number of electrons.
A sphere carrying charge $ne$ has potential:
\[ V = \frac{1}{4\pi\epsilon_0}\,\frac{ne}{r} \]
Put $V = 1.8$, $r = 0.1\ \text{m}$, $\tfrac{1}{4\pi\epsilon_0} = 9\times 10^9$, $e = 1.6\times 10^{-19}$:
\[ 1.8 = 9\times 10^9 \times \frac{n\times 1.6\times 10^{-19}}{0.1} \]
Step 6: Solve for $n$.
Simplify the right side:
\[ 1.8 = 9\times 10^9 \times 1.6\times 10^{-19}\times 10 \times n = 1.44\times 10^{-8}\, n \]
\[ n = \frac{1.8}{1.44\times 10^{-8}} = 1.25\times 10^{8} \]
This matches option (3).
\[ \boxed{n = 1.25\times 10^{8}} \]