Answer
The outward-pointing unit normal is
$\left( {\dfrac{2}{3},\dfrac{2}{3},\dfrac{1}{3}} \right)$
Work Step by Step
At point $P = \left( {2,2,1} \right)$ we have in spherical coordinates:
$\tan \theta = \dfrac{y}{x} = \dfrac{2}{2} = 1$, $\ \ \ \ \ $ $\theta = \dfrac{\pi }{4}$
$\sin \phi = \dfrac{{\sqrt {{x^2} + {y^2}} }}{{\sqrt {{x^2} + {y^2} + {z^2}} }} = \dfrac{{\sqrt {{2^2} + {2^2}} }}{{\sqrt {{2^2} + {2^2} + {1^2}} }} = \dfrac{{\sqrt 8 }}{3} = \dfrac{2}{3}\sqrt 2 $
$\cos \phi = \dfrac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }} = \dfrac{1}{3}$
By Eq. (2):
${\bf{N}} = \left( {{R^2}\sin \phi } \right){{\bf{e}}_r}$
and ${{\bf{e}}_r} = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$, we get the outward-pointing normal at $P = \left( {2,2,1} \right)$:
${\bf{N}} = \left( {9\cdot\dfrac{2}{3}\sqrt 2 } \right)\left( {\cos \dfrac{\pi }{4}\cdot\left( {\dfrac{2}{3}\sqrt 2 } \right),\sin \dfrac{\pi }{4}\cdot\left( {\dfrac{2}{3}\sqrt 2 } \right),\dfrac{1}{3}} \right)$
${\bf{N}} = 6\sqrt 2 \left( {\dfrac{1}{2}\sqrt 2 \cdot\dfrac{2}{3}\sqrt 2 ,\dfrac{1}{2}\sqrt 2 \cdot\dfrac{2}{3}\sqrt 2 ,\dfrac{1}{3}} \right)$
${\bf{N}} = \left( {4\sqrt 2 ,4\sqrt 2 ,2\sqrt 2 } \right)$
So, the outward-pointing unit normal is
$\dfrac{{\bf{N}}}{{||{\bf{N}}||}} = \dfrac{{\left( {4\sqrt 2 ,4\sqrt 2 ,2\sqrt 2 } \right)}}{{\sqrt {32 + 32 + 8} }} = \dfrac{1}{{6\sqrt 2 }}\left( {4\sqrt 2 ,4\sqrt 2 ,2\sqrt 2 } \right)$
$\dfrac{{\bf{N}}}{{||{\bf{N}}||}} = \left( {\dfrac{2}{3},\dfrac{2}{3},\dfrac{1}{3}} \right)$
