1. Let the radius of the spherical balloon at time $t$ be $r(t)$.

2. Surface area of a sphere is $S = 4\pi r^2$ Given that the surface area increases at a constant rate, $\frac{dS}{dt} = \text{constant}$ This means $S$ is a linear function of $t$.

3. Since $S = 4\pi r^2$ we can write $4\pi r^2 = at + b$ for constants $a,b$. Thus, $r^2 = \frac{at+b}{4\pi}$ So $r^2$ varies linearly with time.

4. Use the given data:

• Initially ($t=0$), radius $r=3$ $r^2 = 9$
• After 5 seconds ($t=5$), radius $r=7$ $r^2 = 49$

Since $r^2$ is linear in $t$, let $r^2 = mt + c$ Using $t=0$, $r=3$: $9 = m(0) + c \Rightarrow c=9$ Using $t=5$, $r=7$: $49 = 5m + 9$ $5m = 40$ $m=8$ So, $r^2 = 8t + 9$

5. After 9 seconds: $r^2 = 8(9) + 9 = 72 + 9 = 81$ $r = 9$

6. Therefore, the radius after 9 seconds is $\boxed{9}$

7. Checking options:

• A: $9$ ✅
• B: $10$
• C: $11$
• D: $12$

Hence, the correct option is A.