The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $
So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm | 0)$:

\begin{align*} H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\ \end{align*}

Hint: For the direction, one has to consider the right-hand rule. By this, we see that the $H$-field on the right side points downwards.
Therefore, the sign of the $H$-field is negative.
But here, only the magnitude was questioned!

• for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$
• for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$

• In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
• For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$.
  This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$.
• For $x$ within the outer conductor one also gets a linear proportionality with a similar approach.

The magnitude of the electric displacement field $D$ can be calculated by: $\int D {\rm d}A = Q$.
Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings).
This leads to: \begin{align*} D(x) &= {{Q}\over{A}} \\ &= {{Q}\over{l \cdot 2\pi \cdot x}} \\ \end{align*}

So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm | 0)$:

\begin{align*} D_{\rm i} &= {{Q }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\ D_{\rm o} &= {{Q }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\ \end{align*}

Hint: For the direction, one has to consider the sign of the enclosed charge. By this, we see that the $D$-field is positive.
But here, again only the magnitude was questioned!

• for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$
• for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$

• In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
• Since the charges are on the surface of the conductor, there is no $D$-field within the conductor.