If $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 = 0$ where $a,b,c,d$ are real numbers, what is $a$?

$ 
\textbf{(A)}\ \frac 23
\qquad\textbf{(B)}\ \frac {\sqrt 2} 3
\qquad\textbf{(C)}\ \frac {\sqrt 3} 2
\qquad\textbf{(D)}\ \frac 15
\qquad\textbf{(E)}\ \text{None of the above}
$
Given the equation:
\[ a^2 + b^2 + c^2 + d^2 - ab - bc - cd - d + \frac{2}{5} = 0 \]

We need to find the value of \(a\).

1. **Rewrite the equation in a more manageable form:**
   \[ S = a^2 + b^2 + c^2 + d^2 - ab - bc - cd - d + \frac{2}{5} \]

2. **Complete the square for each variable:**
   - For \(a\) and \(b\):
     \[ a^2 - ab + b^2 = \left(a - \frac{b}{2}\right)^2 + \frac{3b^2}{4} \]
   - For \(b\) and \(c\):
     \[ b^2 - bc + c^2 = \left(b - \frac{c}{2}\right)^2 + \frac{3c^2}{4} \]
   - For \(c\) and \(d\):
     \[ c^2 - cd + d^2 = \left(c - \frac{d}{2}\right)^2 + \frac{3d^2}{4} \]

3. **Substitute these into the original equation:**
   \[ S = \left(a - \frac{b}{2}\right)^2 + \frac{3b^2}{4} + \left(b - \frac{c}{2}\right)^2 + \frac{3c^2}{4} + \left(c - \frac{d}{2}\right)^2 + \frac{3d^2}{4} - d + \frac{2}{5} \]

4. **Combine and simplify the terms:**
   \[ S = \left(a - \frac{b}{2}\right)^2 + \frac{3}{4}\left(b - \frac{2c}{3}\right)^2 + \frac{2}{3}\left(c - \frac{3d}{4}\right)^2 + \frac{5}{8}\left(d - \frac{4}{5}\right)^2 \]

5. **Since \(S = 0\), each squared term must be zero:**
   \[ \left(a - \frac{b}{2}\right)^2 = 0 \implies a = \frac{b}{2} \]
   \[ \frac{3}{4}\left(b - \frac{2c}{3}\right)^2 = 0 \implies b = \frac{2c}{3} \]
   \[ \frac{2}{3}\left(c - \frac{3d}{4}\right)^2 = 0 \implies c = \frac{3d}{4} \]
   \[ \frac{5}{8}\left(d - \frac{4}{5}\right)^2 = 0 \implies d = \frac{4}{5} \]

6. **Substitute back to find \(a\):**
   \[ d = \frac{4}{5} \]
   \[ c = \frac{3d}{4} = \frac{3 \cdot \frac{4}{5}}{4} = \frac{3}{5} \]
   \[ b = \frac{2c}{3} = \frac{2 \cdot \frac{3}{5}}{3} = \frac{2}{5} \]
   \[ a = \frac{b}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5} \]

The final answer is \(\boxed{\frac{1}{5}}\).