A solid brass sphere of radius 21 cm is converted into a right circular cylindrical rod of length 28 cm. The ratio of total surface areas of the rod to sphere is :

Let the radius of the cylindrical rod be $r$.

Volume of the sphere:

$V_{\text {sphere }}=\frac{4}{3} \pi \times (21)^{3}={4} \pi \times 7 \times 21 \times 21$ cu. units

Since volume is conserved:

$V_{\text {cylinder }}=\pi r^{2} \times 28={4} \pi \times 7 \times 21 \times 21$

$r^{2}=\dfrac{4\times 7 \times 21 \times 21}{28}=441$

Thus, $r=\sqrt{441}=21$

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Calculating the surface areas:

Surface area of the sphere:

$S_{\text {sphere }}=4 \pi \times 21^{2}$

Surface area of the cylinder (including the circular ends):

$S_{\text {cylinder }}=2 \pi r h+2 \pi r^{2}$

$=2 \pi \times 21 \times 28+2 \pi \times (21)^{2}$

$= 42\pi(28+21)$

$= 42 \pi \times 49$

Finding the ratio

$\dfrac{S_{\text {cylinder }}}{S_{\text {sphere }}} =\dfrac{42 \pi \times 49}{4 \pi \times (21)^2}$

$=\dfrac{42}{4 \times 9}$

$=\dfrac{7}{2 \times 3}$

$=\dfrac{7}{6}$

Therefore, the ratio of the total surface areas of the rod to the sphere is $\dfrac{7}{6}$.

Hence, the correct answer is Option 2.