The amount of energy carried by a sound wave through a given area in a given time is the intensity $I$. It can be calculated via power $P$ and area $A$ as follows:

$$
begin{align*}
I = frac{P}{A}
end{align*}
$$

For spherical wave expansion the area becomes:

$$
begin{align*}
A = 4 , r^2 , pi
end{align*}
$$

So we get finally:

$$
begin{align*}
I = frac{P}{4 , r^2 , pi} tag{1}
end{align*}
$$

where $r$ is the distance from the sound source to the wave front.

Observers 1 and 3 will experience different sound intensities since they are located at different distances from the sound source. But since it is the same source its power is constant, so we can write formula (1) as follows:

$$
begin{align*}
I_1 &= frac{P}{4 , r_1^2 , pi} \
I_3 &= frac{P}{4 , r_3^2 , pi }
end{align*}
$$

Rearranging into a more useful form we get:

$$
begin{align*}
frac{P}{4 , pi} = I_1 cdot r_1^2 = I_3 cdot r_3^2
end{align*}
$$

We know the intensity measured by the first observer $I_1 = 2.8 cdot 10^{-6} ; frac{text{W}}{text{m}^2}$ , and his distance from the source $r_1 = 1text{ m}$

We also know the intensity measured by the third observer $I_3 = 7.4 cdot 10^{-7} ; frac{text{W}}{text{m}^2}$.

Now rearranging our formula for $r_3$ we get:

$$
begin{align*}
r_3^2 &= frac{I_1 cdot r_1^2}{I_3} \
r_3 &= sqrt{frac{I_1 cdot r_1^2}{I_3}}
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
r_3 = sqrt{frac{2.8 cdot 10^{-6} ; frac{text{W}}{text{m}^2} cdot (1text{ m})^2 }{7.4 cdot 10^{-7} ; frac{text{W}}{text{m}^2} }} approx 1.95text{ m}
end{align*}
$$

$$
I=dfrac{P}{4pi r^2}
$$

Use the equation to solve for P

$$
P=I(4pi r^2)=(3.2times10^{-6};W/m^2)[4pi(48m)^2]=0.093;W
$$

$$
I=dfrac{P}{4pi r^2}
$$

Compute for I in each case.

$I_A=dfrac{10}{4pi (1m)^2}=0.80;W/m^2$

$I_B=dfrac{20}{4pi (2m)^2}=0.40;W/m^2$

$I_C=dfrac{50}{4pi (5m)^2}=0.16;W/m^2$

$I_D=dfrac{100}{4pi(5;m)^2}=0.32;W/m^2$

Hence, the arrangement is

$$
C<D<B<A
$$