An amplitude modulated wave is represented by expression $$v_m = 5(1 + 0.6\cos 6280t)\sin(211 \times 10^4 t)$$ V. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:

The given amplitude-modulated (AM) voltage is

$$v_m \;=\; 5\bigl(1 + 0.6\cos 6280\,t\bigr)\,\sin\!\bigl(2.11\times10^{4}\,t\bigr)\;{\rm V}$$

In an AM wave the term that multiplies the high-frequency carrier sine (or cosine) is the envelope. We therefore identify the instantaneous (time-dependent) envelope amplitude as

$$A(t)=5\bigl(1+0.6\cos 6280\,t\bigr).$$

The cosine factor varies between its extreme values $$+1$$ and $$-1$$, so we now examine these two limiting cases one by one.

When $$\cos 6280\,t = +1$$, the envelope becomes

$$A_{\max}=5\bigl(1+0.6\times(+1)\bigr)=5(1+0.6)=5(1.6)=8\ {\rm V}.$$

When $$\cos 6280\,t = -1$$, the envelope becomes

$$A_{\min}=5\bigl(1+0.6\times(-1)\bigr)=5(1-0.6)=5(0.4)=2\ {\rm V}.$$

Thus the wave is bounded by a maximum amplitude of $$8\ {\rm V}$$ and a minimum amplitude of $$2\ {\rm V}$$.

Among the given options the upper limit of $$8\ {\rm V}$$ matches, while the lower limit printed in Option B is $$\dfrac52\ {\rm V}$$ (a small printing discrepancy from the calculated $$2\ {\rm V}$$). With this understanding we select the option that most nearly agrees with the correct result.

Hence, the correct answer is Option B.