Answer:
1/9 of that just outside the smaller sphere
Step-by-step explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:
![E=k(Q)/(r^2)](https://img.qammunity.org/2020/formulas/physics/high-school/dlapwc2owjbl2muv9l2hawq453hgc50k7g.png)
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is
![E_0=k(Q)/(R^2)](https://img.qammunity.org/2020/formulas/physics/high-school/c5lgpks0m6rvk331qcstsdpe0auyntjy6l.png)
The second sphere has a radius which is 3 times that of the smaller sphere:
![R'=3R](https://img.qammunity.org/2020/formulas/physics/high-school/zmb2xzmmijkxmkdms3uki28j66xe1awckh.png)
So, the electric field just outside the second sphere is
![E'=k(Q)/(R'^2)=k(Q)/((3R)^2)=(1)/(9)(k(Q)/(R^2))=(E_0)/(9)](https://img.qammunity.org/2020/formulas/physics/high-school/i4b8au1vd4o0t0raqd4xe5ge498cfx1v26.png)
So, the correct answer is 1/9.